# Normalization of vacuum state.

1. Jan 10, 2012

### kof9595995

It just occurred to me what if the vacuum state is not normalizable? We usually have the normalization $\langle0|0\rangle=1$, it's acceptable if we are sure the norm of vacuum state is always finite. However, we know states with definite momenta are normalized to delta functions, then how can we be sure vacuum state is not normalized to $\langle0|0\rangle=\delta(0)$?

2. Jan 10, 2012

### alemsalem

I think its because the states with definite momenta are continuous (you have states arbitrarily close in momentum) but the vacuum is isolated from those states.
I think we may have a problem if the mass was zero but maybe not.

3. Jan 10, 2012

### strangerep

Essentially the vacuum is defined that way to be invariant under Poincare transformations,
and to be the lowest-energy eigenstate of the Hamiltonian. That such a vacuum be normalized to 1 is reasonable since if we start with a vacuum and do nothing to it, then a measurement should indicate that we've still got vacuum -- with probability 1.

BTW, delta expressions like $\delta(0)$ don't really make sense. Remember that a Dirac delta is really only a distribution and therefore only makes precise sense when used as a kernel in an integral, e.g.,
$$\int\!dy\; \delta(x-y) \, f(y) ~:=~ f(x)$$
It sometimes helps to think of the delta as $\delta(x,y)$, i.e., having 2 arguments rather than 1. That makes it a bit more obvious that it's a generalization of the Kronecker delta from discrete indices to continuous indices and is really nothing more than an identity mapping.

However, since $\delta(x+a,y+a) = \delta(x,y)$, (i.e., it's translation-invariant), one often takes the shortcut of writing it with only a single argument $x-y$, though this obscures some of its precise meaning at times.

4. Jan 11, 2012

### kof9595995

Well, the same can be said about definite momentum excites states, they are also eigenstates of H.

Yes, but now there is a big qualitative difference between the normalizations of vacuum and excited states, if vacuum isn't that different, then the normalization should be "something like $\delta(0)$".

Last edited: Jan 11, 2012
5. Jan 11, 2012

### kof9595995

Actually I've kept being told a continuum of states indicates nonnormalizability, and surely I haven't found any counter-example to the statement, but I'd really like to see how exactly are they connected.
And I have the same question with you about photon, since I don't recall any special treatment to photon vacuum normalization.

6. Jan 11, 2012

### Bill_K

Sure, this is the issue dealt with in Quantum Electrodynamics under the heading of infrared divergences. Any process may be accompanied by the emission of soft photons which remain undetectable. Thus the photon vacuum is an idealization, and what we really deal with is a continuum of soft photon states.

7. Jan 11, 2012

### alemsalem

I guess the first thing about continuum basis is that you cant just have an ordinary sum, you have to expand states in an integral (if you did the inner product of a generic state would be something like the sum over all numbers in an interval).

if you want to have something like orthonormal basis then <a|a'> = 0 for different vectors but then you can't just have <a|a> = number. because then ∫da' |<a|a'>|^2 = 0.. so you can't have the normalization of the state as just a number (nonnormalizability) and <a|a'> is not an ordinary function its a functional, and actually the integral is not just a limit of a sum of numbers... its just a consistent

8. Jan 11, 2012

### kof9595995

You got me interested, and I found this in a older post(https://www.physicsforums.com/showthread.php?t=468890&page=2):

So what does it mean by " the asymptotic in/out spaces are not Fock spaces"?

9. Jan 11, 2012

### kof9595995

So you mean
$$\int{da'|\langle a|a'\rangle|^2}=\int{da'\langle a|a'\rangle\langle a'|a\rangle}=\langle a|a\rangle$$
So it's contradictory to have <a|a'> = 0 and <a|a> = number at the same time. Seems to be a good point, thanks.

10. Jan 11, 2012

### Bill_K

I believe what the comment means is that a state in Fock space is specified by a definite number of particles occupying each of a finite number of possible states, but in this case it's necessary to consider states having a nonspecific number of soft photons spread over an infinite number of possible states.

11. Jan 12, 2012

### strangerep

The difference is that the vacuum is a physically-realizable state whereas momentum eigenstates are not. We construct more realistic wave-packet states by smearing the momentum eigenstates, e.g.,
$$\def\<{\langle} \def\>{\rangle} \psi_f ~=~ \int\!dp\, f(p) |p\>$$
where f(p) is a Schwarz function. Then
$$\<\psi_g|\psi_f\> ~=~ \int\!dp\, g^*(p) f(p)$$
which is finite because both f,g are Schwartz functions.
http://en.wikipedia.org/wiki/Schwartz_space